a.

\(\left(\frac{11}{12}+\frac{11}{12\times23}+\frac{11}{23\times34}+...+\frac{11}{89\times100}\right)+x=\frac{2}{3}\)

\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{12}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(1-\frac{1}{100}\right)+x=\frac{2}{3}\)

\(\left(\frac{100-1}{100}\right)+x=\frac{2}{3}\)

\(\frac{99}{100}+x=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{99}{100}\)

\(x=\frac{200-297}{300}\)

\(x=-\frac{97}{300}\)

b.

\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{19\times21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\left(\frac{21-11}{231}\right)+\frac{221}{231}-x=\frac{4}{3}\)

\(\frac{10}{231}+\frac{221}{231}-x=\frac{4}{3}\)

\(\frac{231}{231}-x=\frac{4}{3}\)

\(1-x=\frac{4}{3}\)

\(x=1-\frac{4}{3}\)

\(x=\frac{3-4}{3}\)

\(x=-\frac{1}{3}\)

Chúc bạn học tốt

Mình nghĩ câu A phải là 11/23.34 mới đúng.

Theo đề mới: A)

11/12 = 11/1.12

Vậy A = 11/1.12 + 11/12.23 + ... + 11/89.100

= 1 - 1/12 + 1/12 - 1/23 + .... + 1/89 - 1/100

= 99/1000

Vậy x là: 2/3 - 99/100 = -97/300

B) 2/11.13 + ... + 2/19.21 = 1/11 - 1/13 + .... + 1/19 - 1/21 = 10/231

=> 10/231 - x = 4/3 - 221/223 = 229/669

=> x = 10/231 - 229/669

=> x = 6690/154539 - 52899/154539

=> x = -46209/154539 = -15403/51513

Giúp mik luôn bài chứng minh đi bạn T.T

b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)

=>x+99/100=5/3

=>x=5/3-99/100=203/300

c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)

\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)

=>5-x=7/3

hay x=8/3

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

TH : Xét x < -2

Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)

-2x = 2\(\dfrac{1}{4}\)

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

2x + 1,5 = \(\dfrac{3}{4}\)

x = -0,375( loại )

vậy ....

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

b. \(\left(\dfrac{2}{3}-2x\right)\cdot1\dfrac{1}{2}=\dfrac{3}{4}\Rightarrow\dfrac{2}{3}-2x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\Rightarrow-2x=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{6}:\left(-2\right)=\dfrac{1}{12}\)

Vậy \(x=\dfrac{1}{12}\)

c. \(\left|x-1\right|+2\left(x+4\right)=10\Rightarrow\left|x-1\right|+2x+8=10\Rightarrow\left|x-1\right|+2x=10-8=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1+2x=2;x-1\ge0\\-\left(x-1\right)+2x=2;x-1< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2x=2+1+3;x\ge1\\-x+1+2x=2;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=3;x\ge1\\-x+2x=2-1=1;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3:3=1;x\ge1\\x=1;x< 1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)

Vậy x = 1

d. \(\dfrac{11}{12}+\dfrac{11}{12\cdot23}+...+\dfrac{11}{89\cdot100}+x=1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{11}{12}+\dfrac{11}{276}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{22}{23}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{99}{100}+x=\dfrac{5}{3}\Rightarrow x=\dfrac{5}{3}-\dfrac{99}{100}=\dfrac{203}{300}\)

Vậy \(x=\dfrac{203}{300}\)

e. \(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

\(\Rightarrow\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x=\dfrac{7}{3}-4\dfrac{221}{231}\)

\(\Rightarrow x=\dfrac{\dfrac{7}{3}-\dfrac{1145}{231}}{\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}}=\dfrac{-\dfrac{202}{77}}{\dfrac{2}{143}+\dfrac{2}{195}+...+\dfrac{2}{399}}=\dfrac{-\dfrac{202}{77}}{\dfrac{10}{231}}=\dfrac{-202}{77}\cdot\dfrac{231}{10}=\dfrac{-303}{5}\)

Vậy \(x=-\dfrac{303}{5}\)

\(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}-x+\frac{221}{231}=\frac{4}{3}\)

\(=2\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(=2.\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(=\left(\frac{1}{11}-\frac{1}{21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(=\frac{10}{231}-x+\frac{221}{231}=\frac{4}{3}\)

\(\frac{10}{231}-x=\frac{4}{3}-\frac{221}{231}\)

\(\frac{10}{231}-x=\frac{29}{77}\)

\(x=\frac{10}{231}-\frac{29}{77}\)

\(x=\frac{-1}{3}\)

có cần cầu kì vậy ko bn, còn cách nào dễ hỉu hơn ko

a)\(\frac{5}{7}:X=1-\frac{4}{7}=\frac{3}{7}\)

\(X=\frac{5}{7}:\frac{3}{7}=\frac{5}{3}\)

b)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{19}-\frac{1}{21}\right)-X=\frac{4}{3}-\frac{221}{231}=\frac{29}{77}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)-X=\frac{29}{77}\)

\(X=\frac{10}{231}-\frac{29}{77}=-\frac{1}{3}\)

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